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Ch6: the Normal Distribution

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Ch6: the Normal Distribution Ch6: The Normal Distribution Introduction Review: A continuous random variable can assume any value between two endpoints. Many continuous random variables have an approximately normal distribution, which means we can use the distribution of a normal random variable to make statistical inference about the variable. CH6: The Normal Distribution Santorico - Page 175 Example: Heights of randomly selected women CH6: The Normal Distribution Santorico - Page 176 Section 6-1: Properties of a Normal Distribution A normal distribution is a continuous, symmetric, bell-shaped distribution of a variable. The theoretical shape of a normal distribution is given by the mathematical formula (x)2 e 2 2 y , 2 where and are the mean and standard deviations of the probability distribution, respectively. Review: The and are parameters and hence describe the population . CH6: The Normal Distribution Santorico - Page 177 The shape of a normal distribution is fully characterized by its mean and standard deviation . specifies the location of the distribution. specifies the spread/shape of the distribution. CH6: The Normal Distribution Santorico - Page 178 CH6: The Normal Distribution Santorico - Page 179 Properties of the Theoretical Normal Distribution A normal distribution curve is bell-shaped. The mean, median, and mode are equal and are located at the center of the distribution. A normal distribution curve is unimodal. The curve is symmetric about the mean. The curve is continuous (no gaps or holes). The curve never touches the x-axis (just approaches it). The total area under a normal distribution curve is equal to 1.0 or 100%. Review: The Empirical Rule applies (68%-95%-99.7%) CH6: The Normal Distribution Santorico - Page 180 CH6: The Normal Distribution Santorico - Page 181 The Standard Normal Distribution Since each normally distributed variable has its own mean and standard deviation, the shape and location of these curves will vary. To simplify making statistical inference using the normal distribution, we use the standard normal distribution. Standard normal distribution - a normal distribution with mean 0 and a standard deviation of 1. CH6: The Normal Distribution Santorico - Page 182 All normally distributed variables can be transformed into a standard normally distributed variable using the formula for the standard score (z-score): valuemean X z or z . standard deviation The z-score for an observation is the number of standard deviations the observation lies from the mean. The letter Z is used to denote a standard normal random variable. CH6: The Normal Distribution Santorico - Page 183 Example: The mean resting pulse rates for men are normally distributed with a mean of 70 and a standard deviation of 8. A man has a pulse of 80. Find the corresponding z-score. We know: =70, =8, X=80 X 80 70 10 z 1.25 88 The man’s pulse is ____________ standard deviations above the mean because __________ is positive. CH6: The Normal Distribution Santorico - Page 184 Example: The average speed of drivers on I-25 is 63 mph with a standard deviation of 5 mph. Assuming the data are normally distributed, find the z-score for the following speeds clocked by police officers: 74 mph z= 51 mph z= 82 mph z= CH6: The Normal Distribution Santorico - Page 185 Finding Areas/Probabilities for the Standard Normal Distribution Table E in Appendix C gives the area/probability under the standard normal distribution curve to the left of z-values between -3.49 and 3.49 in increments of .01. To find P (Z z) (the probability that a standard normal random variable Z is less than or equal to the value z ), we match our value of z with the left column and top row of Table E (by adding the numbers together). The row and column where these numbers intersect gives us the probability or area under the standard normal curve to the left of z. Note: Use .0001 when z 3.50 and .9999 when z 3.50. CH6: The Normal Distribution Santorico - Page 186 Examples: P (Z 3.49) .0002 P (Z 1.31) .0951 P (Z 0) .5000 P (Z 1.96) .9750 P (Z 3.49) .9998 P (Z 1.43) P (Z 1.82) CH6: The Normal Distribution Santorico - Page 187 Finding the Area Under the Standard Normal Distribution Curve 1. To the left of any z-value: look up the z value in the table and use the area given. CH6: The Normal Distribution Santorico - Page 188 Examples: Find the area to the left of z 0.84, i.e., find P (Z 0.84). Find the area to the left of z = -1.32, i.e., find P (Z 1.32). CH6: The Normal Distribution Santorico - Page 189 2. To the right of any z-value: look up the z value and subtract the area from 1. CH6: The Normal Distribution Santorico - Page 190 Examples: Find the area to the right of z 0.21, i.e., find P (Z 0.21). P(Z > -0.21) = 1 – P(Z -0.21) = 1 – 0.4168 = 0.5832 Find the area to the right of z 1.52, i.e., find P (Z 1.52). CH6: The Normal Distribution Santorico - Page 191 3. Between any two z-values: look up both z values and subtract the smaller area from the larger area. CH6: The Normal Distribution Santorico - Page 192 Examples: Find the area between z 1.71 and z 0.45, i.e., find PZPZPZ( 1.71 0.45) ( 0.45) ( 1.71) 0.6736 - 0.0436 0.63 Find the area between z=1.43 and z=3.01, i.e., find P (1.43 Z 3.01). CH6: The Normal Distribution Santorico - Page 193 Note: For a continuous random variable X (not a discrete random variable), the probability that X equals a specific number is 0. Let c be some number (like 1 or 7). Then P (X c) 0. Intuitive explanation: If we had ten numbers, 1, 2, …, 10, and X is the number we see, P (X 2) 1/10. If we had one thousand numbers, 1, 2, …, 1000, and X is the number we see, P (X 2) 1/1000. For a continuous random variable, there are an infinite number of possibilities, so P (X c) 1/ 0. (This is bad mathematics, but it gets the point across). CH6: The Normal Distribution Santorico - Page 194 Thus, for a continuous random variable: P (Z z) P(Z z) and P (Z z) P(Z z) and P (z1 Z z2 ) P(z1 Z z2 ) P(z1 Z z2 ) P(z1 Z z2 ) CH6: The Normal Distribution Santorico - Page 195 Section 6-2: Applications of the Normal Distribution Finding Probabilities for Normal Distributions To find probabilities for any normal distribution with mean and standard deviation , we simply transform the values on the original scale (call these x, x 1, or x 2) using the formula valuemean x z or z , standard deviation and then use the techniques from the previous section. CH6: The Normal Distribution Santorico - Page 196 Example: An American household generates an average of 28 pounds per month of newspaper for garbage, with a standard deviation of 2 pounds. Assume the amount of newspaper garbage is normally distributed. If a household is selected at random, find the probability of generating between 27 and 31 pounds of newspaper garbage per month. X = garbage. We want P(27 < X < 31). Turn into a Z score so we can use the standard normal distribution. 27X 31 27 28 31 28 PPZ 22 PZ( 0.5 1.5) X PZPZ( 1.5) ( 0.5) Z and 0.9332 0.3085 0.6247 CH6: The Normal Distribution Santorico - Page 197 Example: College women’s heights are normally distributed with 65 inches and 2.7 inches. What is the probability that a randomly selected college woman is 62 inches or shorter? XX 65 Z X=height and 2.7 . X 62 PXP( 62) 62 65 PZ 2.7 We want PZ( -1.11) 0.1335 CH6: The Normal Distribution Santorico - Page 198 Example: What is the probability that a randomly selected college woman is between 63 and 69 inches tall? CH6: The Normal Distribution Santorico - Page 199 Sometimes we need to find the value X that corresponds to a certain percentile. (e.g., what value corresponds to the 98th percentile?) To solve this problem: 1. Find the z that corresponds to this percentile for the standard normal distribution. a. The area to the left of this z-score has the area closest to this percentile. 2. X z . CH6: The Normal Distribution Santorico - Page 200 Example: Mensa is a society of high-IQ people whose members have a score on an IQ test at the 98th percentile or higher. The mean IQ score is 100 with a standard deviation of 16. What IQ score do you need to get into Mensa? 1. First determine what z-score is related to the 98th percentile (to be in the 98th percentile means that 98% of IQ test takers scored at or below your score) (i.e. 0.980). Z = 2.06 2. What is the IQ for the 98th percentile using the formula for X given above? Z = +Z = 100 + 2.06(16) = 132.96 CH6: The Normal Distribution Santorico - Page 201 Example: College women’s heights are normally distributed with a 65 and 2.7. What is the 54th percentile of this population? th 1. Find Z-score for 54 percentile: 2. Find corresponding value for X: CH6: The Normal Distribution Santorico - Page 202 Example: An American household generates an average of 28 pounds per month of newspaper for garbage with a standard deviation of 2 pounds.
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